Generating Function of Bernoulli Polynomials

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Theorem

Let $\map {B_n} x$ denote the $n$th Bernoulli polynomial.

Then the generating function for $B_n$ is:

$\displaystyle \frac {t e^{t x} } {e^t - 1} = \sum_{k \mathop = 0}^\infty \frac {\map {B_k} x} {k!} t^k$


Proof

By definition of the generating function for Bernoulli numbers:

\(\ds \frac t {e^t - 1}\) \(=\) \(\ds \sum_{k \mathop = 0}^\infty \frac {B_k} {k!} t^k\)

By Power Series Expansion for Exponential Function:

\(\ds e^{t x}\) \(=\) \(\ds \sum_{k \mathop = 0}^\infty \frac {x^k} {k!} t^k\)


Thus:

\(\ds \frac {t e^{t x} } {e^t - 1}\) \(=\) \(\ds \paren {\sum_{k \mathop = 0}^\infty \frac {B_k} {k!} t^k} \paren {\sum_{k \mathop = 0}^\infty \frac {x^k} {k!} t^k}\)


Combining like powers of $t$ we obtain:

\(\ds \frac {t e^{t x} } {e^t - 1}\) \(=\) \(\ds \sum_{k \mathop = 0}^\infty t^k \sum_{m \mathop = 0}^k \frac {B_{k - m} } {\paren {k - m}!} \frac {x^m} {m!}\)
\(\ds \) \(=\) \(\ds \sum_{k \mathop = 0}^\infty \frac {t^k} {k!} \sum_{m \mathop = 0}^k \frac {k!} {\paren {k - m}! m!} B_{k - m} x^m\)
\(\ds \) \(=\) \(\ds \sum_{k \mathop = 0}^\infty \frac {t^k} {k!} \sum_{m \mathop = 0}^k \binom k m B_{k - m} x^m\) Definition 1 of Binomial Coefficient
\(\ds \) \(=\) \(\ds \sum_{k \mathop = 0}^\infty \frac {\map {B_k} x} {k!} t^k\) Definition of Bernoulli Polynomial

$\blacksquare$